Air University
Mid Term Examination, Spring-2013
PH-106- Physics 106
Max. Marks: 20
Date: 26-03-2013 Time
Allowed: 2 hrs
Instructors: A. Sadiq, M. Anwar, M. Atif, Rubina Nsar
Note: Attempt all 5 questions. All questions carry equal marks. Please encircle your final
e
answers. You may use basic scientific calculator. DO NOT write in pencil or red
p
ink on answer sheet.
y
p
2
Fig
a
1. Fig 1 shows two protons that are fixed in their places at a distance of 2a  200nm from
1
each other.
a) At what point does the electric field intensity vanish? Where has it only
an x component parallel to the line joining the two protons?
1.0
b) At what point has the electric field intensity has only a y -component perpendicular to
the line joining the two protons? Are there other such points? If so where are these
located?
1.5
c) Make a rough sketch of the electric field intensity as a function of y corresponding
r
P
to part (b). 1.5
O
2. Charge q  0.1C is placed over a conducting sphere of radius r  10cm as shown
in Fig 2.
Fig 2
a) What is its charge density?
0.5
b) What is the electric field intensity E at the point P on at a distance r from its
center O .for, i) r  20cm , ii) r  10cm and (iii) r  5cm
1.5
c) Find the corresponding values of the potential V .
1.5
d) Sketch V as a function of r .
0.5
3. A point charge q  0.1C is placed at a distance r  2cm from the center of a
conducting shell of radius R  20cm .
a) Find the magnitudes and signs of the induced charge on the i) inside and ii) outside
the shell.
1.0
b) Which, if any of these induced charges are uniformly distributed? Explain.
1.0
c) Find the direction and magnitude of the electric field intensity on the outer surface of
the shell.
1.0
d) Sketch of the lines of force near, i) the point charge, ii) the inner and ii) the outer
surface of the shell
1.0
4. A parallel plate capacitor with A  1m 2 has a distance d  0.1mm between its plates
a) What is its capacitance with and without a film of dielectric constant   5 between its
plates?
1.0
b) Find the work done to increase its plate separation to d  0.12mm with initial
1.5
10V across its plates.
c) Sketch the circuit diagram for building C  15 pF capacitor from
three C  10 pF capacitors.
1.5
5. The potential in space in volts varies as V  3x  4 y .
a) What is the magnitude and direction for corresponding electric field intensity at the
origin?
1.5
b) Make a rough sketch of the lines of force corresponding to this potential
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Page 1 of 4
1.5
R
c) Sketch the corresponding equipotential surfaces?
1.0
Solution: Q 1. a) The electric field intensity is zero midway between the two protons. It
is directed away from this point along the x -axis (the line joining the two protons) for all
other points on the line. joining the two protons.
b) The electric field intensity has only a y -components for all points in the plane that is
perpendicular to the line joining the two protons and passes through their midpoint
passing .
E y 
c) Since the electric field intensity vanishes midway between the two protons and it
also vanishes at infinity. It must have a maximum value at some point y  y0
along a line perpendicular to the line joining the two protons. Therefore a ketch
of E  y  as a function of y will look like figure 3.
Fig 3
Q2. a) Since charged placed on a conductor only resides on its surface, we have,

q
10 7

 0.79Cm 2
4r 2 4  3.14  10 2
b) A spherical Gaussian surface passing through a point on and outside the conductor
encloses the entire charge placed on the conductor. Therefore from Gauss’s law we
have,
E

q 
r , r  0.1m
4r 2
This gives, For, i) r  0.2m ,
E


q  9  10 8  10 7 
4
r

r

2
.
25

10
r
NC 1
4r 2
4  10 2
Ii) The point r  0.1m lies on the surface of the conductor,
E


q  9  10 8  10 7 
r
r  8.0  10 4 NC 1 r
2
2
4r
10
iii) r  0.05m . This point lies inside the conductor. A spherical Gaussian surface passing
through a point inside the conductor doesn’t enclose any charge. Therefore from
Gauss’s law we have,
E0
b) The electric potential V r  at a point r in space is given by the work done to bring a
unit positive charge from infinity to that point. In this case for r  0.1 we get,
V r  
q
, r  0.1m . Therefore,
4r 2
For, i) r  0.2m , V 
© Air University, 2013
q
9  10 8  10 7

 450volts
4r
0.2
Page 2 of 4
y
ii) The point r  0.1m lies on the surface of the conductor,
q
9  10 8  10 7
V

 900volts
4r
0.1
iii) The point r  0.05m .lies inside the conductor where the electric field intensity is zero
including at this point. Hence no additional work is done to move a positive charge from
the surface of the conductor to this point. Therefore, we have at this and at every other
point inside the conductor,
V  900volts
Q3. a) The charge induced on the inner surface of the shell is  0.1C and that induced
on its outer surface is 0.1C .
b) The charge induced on the surface of the shell is determined by the strength of the
electric field very closed to it. Also for a conductor the electric field intensity is always
perpendicular to its surface.
From these two considerations we conclude that the charge induced on the inner
surface of the shell is not distributed uniformly as the point charge inside the shell is not
equidistant from all parts of the inner parts of the shell. The induced charge density on
the inner surface of the shell that are closer to the point charge will be higher than on
charge induced on its inner parts that are farther from the point charged.
The charge induced on the outer surface of the shell is uniformly
distributed. This is because the electric field intensity very near the

conducting shell as well as very far from it it is perpendicular to its
surface/
Fig 4
c).Fig 4 shows a sketch of the line force near the point charge and near the
and outer surfaces of the shell.
4/ a) With A  1m 2 , d  0.1mm ,   5
Capacitance without the dielectric material between the plates,
C
0 A
d

8.8  10 12  1
 8.8  10 8 F  88 pF l
4
10
Capacitance with the dielectric material between the plates,
C
 0 A
d

8.8  10 12  1  5
 44  10 8 F  440 pF  0.44F
10 4
b) Charge Q on the plates for a potential difference of 10V across them,
© Air University, 2013
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inner
Q  CV  0.88C
Electric field intensity E between the plates,
E

Q
88  10  8


 10 4 Vm 1
 o A 0 8.8  10 12
Since electric field intensity outside the plates is zero, the average electric force
between the plates is given by,
F
1
1
QE   88  10 8  10 4  44  10  4 N f
2
2
Therefore work don W in increasing the plate separation by
W  Fd  44  10 4  2  10 5  88  10 9 J
c) To get a net capacitance of C  15 pF by joking three 10 pF capacitors we must join
one of the 10 pF in parallel with the two 10 pF capacitors that are joined together in
series. A sketch of the corresponding circuit diagram of the three is shown in
Fig 5.
Fig 5
5. a) For V  3x  4 y V we have,

E   V   x i


y
j


z k

3x  4 y    3i  4


j Vm
1
Magnitude E , E  E x  E y  5Vm 1
2
2
Making an angle  with the x -axis given by,
  tan 1
Ey
Ex
 tan 1
4
 530
3
Since the electric field intensity is constant the above are also it magnitudes and
direction at the origin.
b) Since the electric field corresponding to the given potential is uniform
making an angle 530 with the x -axis, a sketch the projections of its lines of
force in the x, y plane is shown in Fig 6.
.
Fig 6
c) Equipotential surfaces are perpendicular to the lines of force everywhere. In this
case the equipotential surfaces are planes perpendicular to the lines of force shown in
Fig 6. Fig 7 show a project of the equipotential surfaces in the x, y plane. Fig 7
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